Answer:
Explanation:
The crabs cannot see the plankton they eat near the ocean floor. For the crabs to see the plankton, some color of visible light would need to reach the plankton so that it can be reflected into the crabs' eyes.
Answer:
Explanation:
When the amount of H2O2 is doubled while KI is kept constant, the rate of reaction doubles.
When the amount of KI is doubled and the amount of H2O2 is halved, the rate stays nearly constant.
2H2O2 (aq) → O2(g) + 2H2O (l) ------------- first order kinetics reaction.
Catalysts are KI, FeCl3 only, KCl is not a catalyst. Order: KI < MnO2 < Pb < FeCl3.
H2O2 + I– -> IO– + H2O (Step 1)
H2O2 + IO– -> I– + H2O + O2 (Step 2)
It can be seen that the iodine ion (provided by the KI solution) is a product as well as a reactant.
02(g)2Fe? (aq) + 2 H(a) 2 H 2 Fe3 (aq) H2O2(aq) + 2 Fe,Taq) H02(aq) 2 Fe (aq) 2 H (aq)
Answer:
Rate of formation of SO₃
= 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒
-----------------------------(1)
Given that the rate of disappearance of oxygen =
= 3.64 x 10⁻³ M/s
So the rate of formation of SO₃
= ?
from equation (1) we can write
![\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%20%3D%202%20%5B-%5Cfrac%7Bd%5BO_%7B2%7D%5D%20%7D%7Bdt%7D%20%5D)
⇒
= 2 x 3.64 x 10⁻³ M/s
⇒
= 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃
= 7.28 x 10⁻³ M/s
Answer:
Explanation:
The reaction is given as:

The reaction quotient is:
![Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Q_C%20%3D%20%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
From the given information:
TO find each entity in the reaction quotient, we have:
![[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%20%5Cdfrac%7B6.42%20%5Ctimes%2010%5E%7B-4%7D%7D%7B3.5%7D%5C%5C%20%5C%5C%20NH_3%20%3D%201.834%20%5Ctimes%2010%5E%7B-4%7D)
![[N_2] = \dfrac{0.024 }{3.5}](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%5Cdfrac%7B0.024%20%7D%7B3.5%7D)
![[N_2] = 0.006857](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%200.006857)
![[H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%5Cdfrac%7B3.21%20%5Ctimes%2010%5E%7B-2%7D%7D%7B3.5%7D)
![[H_2] = 9.17 \times 10^{-3}](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%209.17%20%5Ctimes%2010%5E%7B-3%7D)
∴

However; given that:

By relating
, we will realize that 
The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.